Respuesta :
Given that,
Sample size= 83
Mean number= 39.04
Standard deviation= 11.51
We know the critical t-value for 95% confidence interval which is equal to 1.989.
We also know the formula for confidence interval,
CI=( mean number - critical t-value*standard deviation/(sample size)^(1/2), mean number + critical t-value*standard deviation/(sample size)^(1/2))
So, we have
CI= (39.04 - 1.989*11.51/83^(1/2), 39.04 + 1.989*11.51/83^(1/2)
CI= (39.04 - 2.513,39.04 + 2.513)
CI= (36.527,41.553)
Therefore, 95% confidence interval for these data is (36.527,41.553), and this result interpret that the true value for this survey sample lie in the interval (36.527,41.553).
Answer: We are 95% confidence that the respondents are worked between 36.57 hours and 41.51 hours.
Explanation:
Since we have given that
Number of respondents in a survey sample = 83
Mean number of hours worked per week = 39.04
Standard deviation = 11.51
Now, we'll calculate standard error of mean is given by
[tex]\sigma_\bar{x}=\frac{\sigma}{\sqrt{n}}=\frac{11.51}{\sqrt{83}}=1.26[/tex]
95% of confidence interval is given by
[tex]\bar{x}\pm 1.96\sigma_{\bar{x}}\\\\39.04\pm 1.96\times 1.26\\\\39.04\pm 2.47\\\\\{41.51,36.57\}[/tex]
Hence, we are 95% confidence that the respondents are worked between 36.57 hours and 41.51 hours.
